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Dual of electromagnetic field tensor.

  1. Jun 14, 2008 #1
    What is the importance of dual electromagnetic field tensor? Generally this is not included in the action. What will be the advantage/disadvantage if I include terms like
    F_{\mu \nu}\tilde{F}^{\mu \nu}, \tilde{F}_{\mu \nu}\tilde{F}^{\mu \nu}
    in the action? (The tilde denotes the dual tensor.)
    Also does the property of dual tensor depend on the spacetime dimension?

    It seems ultimately these can be simplified and written in terms of [tex]F_{\mu \nu}[/tex] alone but the [tex]\epsilon^{\mu \nu \rho \sigma} [/tex] will give rise to new features. These I expect will depend on the dimension. But some interesting new features should also emerge. I am thinking about it. Meanwhile you may help with your comments. Thanks.
    Last edited: Jun 14, 2008
  2. jcsd
  3. Jun 14, 2008 #2
    The contraction of the electromagnetic field strength tensor with its dual, [itex]F^{\mu\nu}\tilde{F}_{\mu\nu}[/itex], as it turns out, can be written as a total derivative after integrating that term by parts (and assuming field configurations fall off quickly enough at infinity). Then by gauss' law, that term will not contribute to the dynamics, and for this reason is dropped.
    Exercise: show this.

    The contraction of the dual with another dual is merely proportional to the ordinary gauge kinetic term with the strength tensor contracted with itself, i.e. [itex]\tilde{F}^{\mu\nu}\tilde{F}_{\mu\nu}\propto F_{\mu\nu}F^{\mu\nu}[/itex]. And for this reason, it would be redundant to include this in the action.
    Exercise: show this also.

    Please note that in nonabelian gauge theories, there could be field configurations for which the fields don't fall off fast enough at infinity. In this case, the surface term will introduce new physics.
  4. Jun 15, 2008 #3
    For more on what TriTertButoxy (wtf username is that? =), you can look at Chern-Simons theories, also instanton density. Also, the dual and the field are equivalent in 4D provided there are no sources. This is the so-called electromagnetic duality. Finally, a lot of important topological features of the gauge theory are encoded in so-called characteristic classes, and the dual shows up there also.
  5. Jun 17, 2008 #4
    Your first example is the so-called [tex]\theta[/tex]-term, which can be written as a total derivative. The second example can be directly related to usual kinetic term for gauge field. Due to [tex]\epsilon[/tex], Levi Civita tensor, the P and CP would be violated.
    Since there is no observation of CP violation, so in QED the [tex]\theta[/tex]-term is hence forbidden.
    As for the non-Abelian gauge theory, the [tex]F_{\mu\nu}\tilde{F}^{\mu\nu}[/tex] can also be written as a total derivative, and the current under derivative is the Chern-Simons current. (It happened to be one of my homework for particle physics course :P)
    But this time the field may not go off sufficiently rapid such that the [tex]\theta[/tex]-term matters. This is the strong CP problem.
    However, I'm not very clear about the instanton and the topology of QCD vacuum, is there some introductory article? (I searched some papers, but full of differential forms...lots of mathematics..) Thanks for any references and suggestions.
  6. Jun 17, 2008 #5


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    Track down Wittens papers in the 80s for reviews of that sort of thing.

    You will get differential forms though no matter where you look, b/c thats sort of the natural way to proceed. However as I recall, in one of Wittens papers he summarizes and explains their usage in a really clear and elegant way so that it all sort of flows naturally.

    Also I think T'Hooft has some introductory notes on the subject as well (maybe in his QFT book) that I had used when I was learning.
  7. Jun 17, 2008 #6


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  8. Jun 18, 2008 #7
    Sorry but I can't help you there. However I can tell you that [itex]\tilde{F}_{\mu \nu}\tilde{F}^{\mu \nu}[/itex] is a term found in the expression for the stress-energy-momentum tensor of the electromagnetic field.

  9. Jun 18, 2008 #8
    For instantons in QCD, I would strongly suggest to begin here :
    Instantons at work

    Also, I think Haelfix was refering to
    Monopoles, Instantons and Confinement
    Last edited: Jun 18, 2008
  10. Jun 19, 2008 #9

    Hans de Vries

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    It shows up in the Gordon decomposition of the axial vector current. The part of the
    axial current dependent on the phase change rates can for instance be written as:

    j_{A\phi}\ =\ \frac{e}{\epsilon_o}
    \left[~ \begin{array}{cccc}
    0 & - c\tilde{M}_x & - c\tilde{M}_y & - c\tilde{M}_z \\
    ~~c\tilde{M}_x & 0 & ~~~P_z & - ~P_y \\
    ~~c\tilde{M}_y & - ~P_z & 0 & ~~~P_x \\
    ~~c\tilde{M}_z & ~~~P_y & - ~P_x & 0
    \left[~ \begin{array}{c}
    \,~~j^o \\
    \,- j^x \\
    \,- j^y \\
    \,- j^z

    Where M and P are the magnetization and polarization of the Dirac wavefunction as
    described by Ohanian in: What is Spin?—[American Journal of Physics 54, 500 (1986)]

    And j at the RHS is the convection current density part of the Gordon decomposition of the
    vector current density. This is the part which is the same as the Klein Gordon current.

    It also shows up in the covariant definition of the Spin of the electromagnetic field,
    The Chern Simons current.

    {\cal S}_\mu\ =\
    \begin{array}{c c c c}
    ~ 0 & -c\,\textsf{H}_x & -c\,\textsf{H}_y & -c\,\textsf{H}_z \\
    c\,\textsf{H}_x & ~~~ 0 & \ \ ~~\textsf{D}_z & ~-\textsf{D}_y \\
    c\,\textsf{H}_y & ~-\textsf{D}_z & ~~~ 0 & \ \ ~~\textsf{D}_x \\
    c\,\textsf{H}_z & \ \ ~~\textsf{D}_y & ~-\textsf{D}_x & ~~~ 0
    \right] \times \left[
    \ \ A_0 \\
    -A_x \\
    -A_y \\

    Where the A at the RHS have to be the Lienard Wiechert potentials.

    Regards, Hans
    Last edited: Jun 19, 2008
  11. Jun 20, 2008 #10
    Thanks everybody for your responses. In fact I am yet to go through all the references and digest the relevant points. however here I point out what I have found while trying the exercises suggested by TriTertButoxy.
    1. Using the fact that
    \tilde{F}^{\mu \nu}=\frac{1}{2}\epsilon^{\alpha \beta \mu \nu}F_{\alpha \beta}
    I have
    \int d^4x F_{\mu \nu}\tilde{F}^{\mu \nu}=-\int d^4x \partial_{\mu}\left( \epsilon^{\mu \nu \alpha \beta}F_{\alpha \beta}A_{\nu} \right)
    This is not a total derivative. However this will be a total derivative only if
    \epsilon^{\mu \nu \alpha \beta}F_{\alpha \beta}\partial_{\mu}A_{\nu}=0
    which is not true. I hope I have not made mistake or overlooked something. Can anybody help me with this?
    2. I have shown that in four dimension,
    \tilde{F}^{\mu\nu}\tilde{F}_{\mu\nu} = F_{\mu\nu}F^{\mu\nu}
    Hope this is correct. If not then there should be some numerical factors. Can anybody tell if there is any such factor. I have done in hurry, so some minor mistake is possible.
  12. Jun 20, 2008 #11
    In the previous post I made a mistake in the 1st eqn, it should be:
    \int d^4x F_{\mu \nu}\tilde{F}^{\mu \nu}=-\int d^4x \partial_{\mu}\left( \epsilon^{\mu \nu \alpha \beta}F_{\alpha \beta} \right)A_{\nu}
    Others comments are unchanged.
  13. Jun 30, 2008 #12
    Hi, I'm posting solutions to the exercises:

    Exercise 1) Starting with the definition of the field strength tensor and its dual, we have:

    [tex]\begin{array}{rcl}\displaystyle \tilde F_{\mu\nu}F^{\mu\nu}&=&\displaystyle\epsilon_{\mu\nu\rho\sigma}(\partial^\rho A^\sigma-\partial^\sigma A^\rho)(\partial^\mu A^\nu-\partial^\nu A^\mu)\\
    &=&\displaystyle4\epsilon_{\mu\nu\rho\sigma}(\partial^\rho A^\sigma)(\partial^\mu A^\nu)\\
    &=&4\epsilon_{\mu\nu\rho\sigma}(\partial^\mu \partial^\rho A^\sigma)A^\nu-4\epsilon_{\mu\nu\rho\sigma}\partial^\mu(\partial^\rho A^\sigma A^\nu),\end{array}[/tex]​

    where symmetry projection rules were used in going from the first line to the second. Notice that the [itex](\partial^\mu \partial^\rho A^\sigma)[/itex] factor in the first term is symmetric in [itex]\mu[/itex] and [itex]\rho[/itex] indices because derivatives commute. But since it is contracted into the totally antisymmetric tensor, the first term vanishes. The second term is the total derivative, which can be dropped as long as fields drop off quickly enough at infinity.

    Exercise 2) Again, start with the definition of the field strength dual.

    [tex]\begin{array}{rcl}\displaystyle \tilde F_{\mu\nu}\tilde F^{\mu\nu}&=&\displaystyle\epsilon_{\mu\nu\rho\sigma}\epsilon^{\mu\nu}_{\phantom{\mu\nu}\alpha\beta}F^{\rho\sigma}F^{\alpha\beta}\\

    Note the ordering of indices in the second term of the last line. This proves that the contraction of the dual field strength tensor with another dual tensor is proportial to the ordinary gauge kinetic term.

    I hope that helps!
  14. Sep 3, 2008 #13


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    I am storming this old thread because it happens to have some references I was going to give anyway. I want to talk on monopoles, instantons and EM duality.

    To give an start, let me tell that the two papers of dirac are scanned online by a third party source:
    Quantised Singularities in the Electromagnetic Field, Proc. Roy. Soc. A 133, p. 610-624, (1931) http://home.tiscali.nl/physis/HistoricPaper/Dirac/Dirac1931.pdf
    The Theory of Magnetic Poles, Phys. Rev. 74, p. 817-830, (1948) http://home.tiscali.nl/physis/HistoricPaper/Dirac/Dirac1948.pdf

    My first question is: [itex]g e = n[/itex], or [itex]g e=n/2[/itex]??? The paper of 1948 says n/2, but I think that in more recent work it is said that n must be even.

    hep-th/9506077 uses [itex]e g= 2 \pi[/itex] with, they say, [tex]\hbar =1[/tex]
    hep-th/9509066 uses [itex]e g \propto 1[/itex]
    hep-th/9407087 hep-th/9408099 are unclear about it, or do not mention it.
    Last edited: Sep 3, 2008
  15. Sep 5, 2008 #14


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    A nice review of monopoles up to Seiberg Witten theory is Ketov's hep-th/9611209v3

    There we can see (pg 20) that if we apply the DSZ (or DZS or DSZW) quantisation to the n_e=1 BPS solution (of Giorgi model, as usual) we have

    e + g = \sqrt{4 \pi} (\sinh ^{-1/2} \gamma + \sinh ^{+1/2} \gamma)

    With gamma a free parameter in the classical solution.

    Hans "1st order" observation, on the other hand, tell us that in Nature, with usual electromagnetism

    e + g = \sqrt{4 \pi} (\sinh {\pi^2 \over 4} + \cosh {\pi^2 \over 4} )

    so that 1) the shape involves exponential but not the same expresion and 2) the internat hiperbolic parameter is fixed.

    I wonder, about (1), if other models of BPS monopoles can produce this shape instead of Giorgi's BPS.
  16. Sep 5, 2008 #15


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    Shnir uses a different parametrization for the free parameter the BPS solution. From 5.56 and 5.64ff (pg 126) it seems that the equivalence is
    \tan \alpha \to \sinh \gamma
    making more of the point of obtaining the vacuum of this solution as a rotation of the one of the plain monopole.

    The book has also some sparse comments on the relation between the monopole/dyon solutions and those of instantons. I am intrigued because, as said elsewhere, exp(pi^2/4) seems very much as wave packet rotated to Euclidean space. I wished to know more about how the Euclidean solutions are normalised.
  17. Sep 6, 2008 #16


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    Why do I think that Hans approximation looks as an instanton? Yes, it is true that textbooks are full with expresions of the kind exp(4 pi^2 /g), but such g is the coupling constant, not an exact number. On the other hand, if one goes to integral tables it is easy to find a rewriting

    e + g = \sqrt{4 \pi} \; {\bf e}^{\pi^2 \over 4} = \int {\bf e}^{\pm{\pi \over 2}} {\bf e}^{-{x^2 \over 4}} dx
    or more generally, for any [itex]k[/itex]
    e + g = 2 k \int_{-\infty}^{\infty} {\bf e}^{\pm k \pi} {\bf e}^{-{k^2 x^2}} dx

    Such integrals seem very much as gaussian wavepackets with momentum [itex]k_0= ik\pi[/itex], thus imaginary momentum, thus wick-rotated into Euclidean.

    My problem is that the normalisation does not fit so easily, or I do not understand how an instanton solution should be normalised.

    the implied eigenfunction, rotated back, and rootsquared, could be perhaps

    F= \sqrt{2 k} {\bf e}^{ \pm i k \pi /2} {\bf e}^{-{k^2 x^2}/2} dx

    so that <F|F*> is again
    2 k \int_{-\infty}^{\infty} {\bf e}^{-{k^2 x^2}} dx = \sqrt {4 \pi}

    but there are other alternative hand wavings. For instance, I could just rotate, without rootsquaring, and then I would integrate to
    4 k^2 \int_{-\infty}^{\infty} {\bf e}^{-{2 k^2 x^2}} dx = k \sqrt 2 \sqrt { 4 \pi}

    Also it is not to be discarded a 3D radial spherical integration, instead of a lineal (which has not very sense, anyway, if it is not along time). Then extra factors 4 pi will come.
    Last edited: Sep 6, 2008
  18. Sep 6, 2008 #17


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    hmm errata, of course in the previous message the exponentials contain k x, not simply k.
  19. Sep 9, 2008 #18


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    Ok, given the disaster of the errata, let me to explain again what is my issue with the normalisation of instantons. Lets put from any textbook. I am using page 45 of Bars.

    [tex]|\Psi(x)|^2={ 1 \over \sqrt {2 \pi a} } e^{x^2 \over 2 a} [/tex]

    this is the probability distribution of a wave packet (note the freedom of p_0)
    [tex]\Psi(x)={ 1 \over \sqrt {\sqrt {2 \pi a} } } e^{i p_0 x \over \hbar} e^{x^2 \over 4 a} [/tex]

    of amplitudes
    [tex]\Delta x = \sqrt a[/tex] and [tex] \Delta p = {\hbar \over 2 \sqrt a} [/tex]

    and of course [itex] \int |\Psi(x)|^2 dx =1[/itex]

    Now, consider the instanton, or Euclidean solution
    [tex]\Psi_E(x)={ 1 \over \sqrt {\sqrt {2 \pi a } } } e^{ p_0 x \over \hbar} e^{x^2 \over 4 a} [/tex]

    its density or probability distribution is now
    [tex]|\Psi_E(x)|^2= {1 \over \sqrt {2 \pi a} } e^{2 p_0 x \over \hbar} e^{x^2 \over 2 a} [/tex]
    and its total sum gets an extra factor
    [tex]N_E \equiv \int |\Psi_E(x)|^2 dx = e^{ 2 a p_0^2 \over \hbar^2} [/tex]
    depending of the "momentum boost" [itex]p_0[/itex].

    Our first position to confront such discrepancy is to postulate that the instanton must be pointlike (thus [itex]a=0[/itex]) with p_0 finite or perhaps static, thus [itex]p_0=0[/itex] for some finite width [itex]a[/itex]. Is it reasonable? It is not, because quantum mechanics allows for an indeterminacy of order [itex]\propto \Delta p[/itex]

    A second, more reasonable, position, is to say that a quantum instanton could have a boost bounded under the cloak of the indeterminacy principle. This could be
    [tex] p_0 \propto \Delta p = {\hbar \over 2 \sqrt a} [/tex]
    and then
    [tex]N_E \approx e^{1/2} [/tex]

    This is "almost" the condition to get Hans' formula. We could get it by using some alternate undeterminacy bound or some alternate introduction of the p0 bost.

    But perhaps it is not the exact principle to use. For instance, a Gedaken experiment next in the same book gets
    [tex] \Delta p \Delta x > 4 \pi \hbar = 2 \bf {h} [/tex]

    and then using
    [tex] p_0 \propto \Delta p = {2 \bf {h} \over{\sqrt a} } [/tex]
    we get
    [tex]N_E \approx e^{32 \pi^2} [/tex]

    So I guess I am near of Hans factor, but not right on target.
  20. Mar 17, 2009 #19
    hello, i am struggling to do this same exercise with the non-abelian (yang-mills) field strength tensor. can anyone help me?

    [tex]\begin{array}{rcl}\displaystyle F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} - i [A_{\mu}, A_{\nu}] \end{array}[/tex]​
  21. Mar 19, 2009 #20


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    1) Use

    F^{*}_{ab} = \frac{1}{2} \epsilon_{abcd} F^{cd} = \epsilon_{abcd} (\partial^{c}A^{d} + A^{c}A^{d} ) \ \ \ (1)

    to show that

    D^{a} F^{*}_{ab} \equiv \partial^{a} F^{*}_{ab} + [ A^{a} , F^{*}_{ab} ] = 0 \ \ \ (2)

    (this follows simply from expanding out the terms and using the antisymmetry of [itex]\epsilon_{abcd}[/itex])
    Notice that this is an off-shell identity. It does not require [itex]F_{ab}[/itex] to be a solution of the Yang-Mills equation [itex]D^{a}F_{ab} = 0[/itex].

    2) write

    \begin{array}{rcl}\displaystyle Q &=& \displaystyle Tr \{ F^{ab}F^{*}_{ab} \} \\
    &=&\displaystyle Tr \{ \partial^{[a}A^{b]} F^{*}_{ab} + A^{[a}A^{b]} F^{*}_{ab}\} \\
    &=& \displaystyle Tr \{ \partial^{[a}A^{b]}F^{*}_{ab} + A^{a}[A^{b},F^{*}_{ab}]\} \\
    &=& \displaystyle Tr \{ \partial^{[a}A^{b]}F^{*}_{ab} - A^{a} \partial^{b}F^{*}_{ab}\} \\
    &=& Tr \{ \partial^{a}A^{b}F^{*}_{ab} - \partial^{b} ( A^{a}F^{*}_{ab} ) \} \end{array}

    The 3rd line follows from the cyclic property of the trace and the 4th line comes from eq(2).

    3) expand out [itex]F^{*}_{ab}[/itex] as in eq(1) and use the identity

    Tr \{ \epsilon_{abcd} ( \partial^{a} A^{b}) A^{c}A^{d} \} = \frac{1}{3} Tr \{ \epsilon_{abcd} \partial^{a}(A^{b}A^{c}A^{d}) \}

    which again follows from the cyclicity of the trace and the antisymmetry of [itex]\epsilon_{abcd}[/itex].

    4) you get

    [tex] Q = \partial^{a} J_{a}[/tex]

    [tex] J_{a} = \epsilon_{abcd} Tr \{ A^{b} ( \partial^{c} A^{d} + \frac{2}{3} A^{c}A^{d} ) \}[/tex]


    Last edited: Mar 19, 2009
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